Math 214 - Multivariable Calculus - Prep

Task 37.1

Consider the region that lies below the $z$-axis and between 2 spheres of radii $a$ and $b$ with $a<b$. The image below shows such a region where the inner radius is $a=3$, and the outer radius is $b=5$.
1. Set up an iterated triple integral in spherical coordinates to compute the volume of this region.
2. Set up an iterated triple integral formula to compute the $z$-coordinate of the centroid (symmetry gives us $\bar x = \bar y = 0$).
3. If the temperature at points in this region is given by $T(x,y,z) = x+3$, then set up an iterated triple integral formula that would give the average temperature of the region.
A metal casing lies inside the cylinder $x^2+y^2=4$, outside the cylinder $x^2+y^2=1$, below the paraboloid $z=9-x^2-y^2$, and above the plane $z=0$. The region is shown below, where one quarter of the region was removed so you can see the hollow interior.
1. Set up an iterated integral in cylindrical coordinates to compute the volume of the casing.
2. The casing is made of a composite material and the density of the casing is more dense the further from the center. The density is given by $\delta(x,y,z) = x^2+y^2$. Set up an iterated triple integral formula to compute the $z$-coordintate of the center-of-mass of the casing.

Remember that you can verify that your bound are correct by using Mathematica to draw whatever you decide the bounds should be. Here's two examples of how to construct regions, the first in cylindrincal coordinates, and the second in spherical coordinates.

coordinates = {r Cos[theta], r Sin[theta], z}
R = ParametricRegion[coordinates, {{r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[r, {r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}]

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
R = ParametricRegion[coordinates, {{theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}}];
Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}]
Integrate[rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}]

If the code above is extremely slow on your computer, then you can use the code below instead. The code is more complicated, as it plots the six surfaces defined by the bounds you choose, but the code used requires very minimal processing (it works fast). (The Evaluate[] command is needed for the code to work prior to Mathematica 13.)

plotRegion3D[cs_, ob_, mb_, ib_] := 
 Show[{{ParametricPlot3D[Evaluate[Table[cs /. (ib[[1]] -> ib[[i]]), {i, 2, 3}], ob, mb], AxesLabel -> {x, y, z}, Mesh -> {15, 1}], 
    ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> mb[[i]]), {i, 2, 3}], ob], {s, 0, 1}, PlotStyle -> {Red, Blue}, Mesh -> {15, 0}], 
    ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> (mb[[2]] (1 - t) + mb[[3]] t)) /. (ob[[1]] -> ob[[i]]), {i, 2, 3}]], {t, 0, 1}, {s, 0, 1}, PlotStyle -> Green, Mesh -> {0, 0}]}}, PlotRange -> All]

coordinates = {r Cos[theta], r Sin[theta], z}
plotRegion3D[coordinates, {r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}]

coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]}
plotRegion3D[coordinates, {theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}]

Much simpler code will draw 2D regions

plotRegion[cs_, ob_, ib_] := ParametricPlot[Evaluate[cs, ob, ib], Mesh -> {10, 0}]

coordinates = {r Cos[theta], r Sin[theta]};
plotRegion[coordinates, {theta, Pi/4, Pi}, {r, 2, 5}]

Task 37.2

Consider the region $R$ in space satisfying $0\leq x-y\leq 4$ and $1\leq 2x+y\leq 3$. We wish to evaluate the integral $\ds\iint_R xy dA$.
1. Draw the region $R$.
2. Using the change-of-coordinates $u=x-y$ and $v=2x+y$, compute the Jacobian $\frac{\partial(u,v)}{\partial (x,y)}$.
3. Find $x$ and $y$ in terms of $u$ and $v$.
4. Use this change-of-coordinates to compute $\ds\iint_R xy dA$ by first setting up an appropriate iterated integral of the form $\ds \int_{?}^{?}\int_{?}^{?}?dudv$.
Consider the ellipsoid $\ds\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, for some positive constants $a$, $b$, and $c$.
1. Draw the region.
2. Using the change-of-coordinates $x = a u \sin v \cos w$, $y = b u \sin v \sin w$, $x = c u \cos v$, compute the Jacobian $\frac{\partial (x,y,z)}{\partial (u,v,w)}$. Feel free to use software to help you.
3. Set up an iterated integral using $uvw$-coordinates to compute the volume inside the ellipsoid. Then compute the integral.
4. Set up an iterated integral using $uvw$-coordinates to show that $\bar z = \frac{3c}{16}$ for the region inside the ellipsoid that is above the $xy$-plane.

Task 37.3

When you can use a potential to compute work, it greatly simplifies things.

As you complete each problem below, first ask if there is a potential.
1. Compute the work done by the vector field $\vec F(x,y) = (y,x)$ on a object that moves along the path $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2 \pi$.
2. Compute the work done by the vector field $\vec F(x,y) = (-y,x)$ on a object that moves along the path $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2 \pi$.
3. For each vector field above, use Mathematica to construct an image that shows the vector field along with the curve in the same plot. (You'll need VectorPlot[] and ParametricPlot[], with Show[] to get them in the same plot).

We've seen that if a vector field has a potential, then the derivative is symmetric. Is the converse of this statement true, namely if the derivative of a vector field is symmetric, then does that mean the vector field has a potential?

Consider the vector field $\ds\vec F(x,y) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.
1. Show that the derivative is symmetric.
2. Compute the work done by the vector field $\vec F(x,y)$ on a object that moves along the path $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2 \pi$.
3. Explain why $\vec F(x,y)$ does not have a potential.
4. Look up "simply connected region" (see section 6.3), and explain why the domain of $\ds\vec F(x,y) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$ is NOT simply connected.

When the domain of a continuously differentiable vector field is simply connected, then the vector field has a potential if and only if the derivative is symmetric. The concept of a simply connected domain is the start of an entire branch of mathematics called algebraic topology, all stemming from the question, "under what circumstances can we guarantee that a vector field will have a potential?"

Task 37.4

Pick some problems related to the topics we are discussing from the Text Book Practice page.

Big View
Prep Day 37 Task 37.1 Consider the region that lies below the $z$-axis and between 2 spheres of radii $a$ and $b$ with $a<b$. The image below shows such a region where the inner radius is $a=3$, and the outer radius is $b=5$. Set up an iterated triple integral in spherical coordinates to compute the volume of this region. Set up an iterated triple integral formula to compute the $z$-coordinate of the centroid (symmetry gives us $\bar x = \bar y = 0$). If the temperature at points in this region is given by $T(x,y,z) = x+3$, then set up an iterated triple integral formula that would give the average temperature of the region. A metal casing lies inside the cylinder $x^2+y^2=4$, outside the cylinder $x^2+y^2=1$, below the paraboloid $z=9-x^2-y^2$, and above the plane $z=0$. The region is shown below, where one quarter of the region was removed so you can see the hollow interior. Set up an iterated integral in cylindrical coordinates to compute the volume of the casing. The casing is made of a composite material and the density of the casing is more dense the further from the center. The density is given by $\delta(x,y,z) = x^2+y^2$. Set up an iterated triple integral formula to compute the $z$-coordintate of the center-of-mass of the casing. Remember that you can verify that your bound are correct by using Mathematica to draw whatever you decide the bounds should be. Here's two examples of how to construct regions, the first in cylindrincal coordinates, and the second in spherical coordinates. coordinates = {r Cos[theta], r Sin[theta], z} R = ParametricRegion[coordinates, {{r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}}]; Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}] Integrate[r, {r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}] coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]} R = ParametricRegion[coordinates, {{theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}}]; Region[R, Axes -> True, AxesLabel -> {x, y, z}, AxesOrigin -> {0, 0, 0}] Integrate[rho^2 Sin[phi], {theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}] If the code above is extremely slow on your computer, then you can use the code below instead. The code is more complicated, as it plots the six surfaces defined by the bounds you choose, but the code used requires very minimal processing (it works fast). (The Evaluate[] command is needed for the code to work prior to Mathematica 13.) plotRegion3D[cs_, ob_, mb_, ib_] := Show[{{ParametricPlot3D[Evaluate[Table[cs /. (ib[[1]] -> ib[[i]]), {i, 2, 3}], ob, mb], AxesLabel -> {x, y, z}, Mesh -> {15, 1}], ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> mb[[i]]), {i, 2, 3}], ob], {s, 0, 1}, PlotStyle -> {Red, Blue}, Mesh -> {15, 0}], ParametricPlot3D[Evaluate[Table[(cs /. (ib[[1]] -> (ib[[2]] (1 - s) + ib[[3]] s))) /. (mb[[1]] -> (mb[[2]] (1 - t) + mb[[3]] t)) /. (ob[[1]] -> ob[[i]]), {i, 2, 3}]], {t, 0, 1}, {s, 0, 1}, PlotStyle -> Green, Mesh -> {0, 0}]}}, PlotRange -> All] coordinates = {r Cos[theta], r Sin[theta], z} plotRegion3D[coordinates, {r, 1, 3}, {theta, Pi/2, 2 Pi}, {z, 0, r}] coordinates = {rho Sin[phi] Cos[theta], rho Sin[phi] Sin[theta], rho Cos[phi]} plotRegion3D[coordinates, {theta, 0, Pi/2}, {phi, Pi/6, Pi/3}, {rho, 1, 3}] Much simpler code will draw 2D regions plotRegion[cs_, ob_, ib_] := ParametricPlot[Evaluate[cs, ob, ib], Mesh -> {10, 0}] coordinates = {r Cos[theta], r Sin[theta]}; plotRegion[coordinates, {theta, Pi/4, Pi}, {r, 2, 5}] Task 37.2 Consider the region $R$ in space satisfying $0\leq x-y\leq 4$ and $1\leq 2x+y\leq 3$. We wish to evaluate the integral $\ds\iint_R xy dA$. Draw the region $R$. Using the change-of-coordinates $u=x-y$ and $v=2x+y$, compute the Jacobian $\frac{\partial(u,v)}{\partial (x,y)}$. Find $x$ and $y$ in terms of $u$ and $v$. Use this change-of-coordinates to compute $\ds\iint_R xy dA$ by first setting up an appropriate iterated integral of the form $\ds \int_{?}^{?}\int_{?}^{?}?dudv$. Consider the ellipsoid $\ds\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, for some positive constants $a$, $b$, and $c$. Draw the region. Using the change-of-coordinates $x = a u \sin v \cos w$, $y = b u \sin v \sin w$, $x = c u \cos v$, compute the Jacobian $\frac{\partial (x,y,z)}{\partial (u,v,w)}$. Feel free to use software to help you. Set up an iterated integral using $uvw$-coordinates to compute the volume inside the ellipsoid. Then compute the integral. Set up an iterated integral using $uvw$-coordinates to show that $\bar z = \frac{3c}{16}$ for the region inside the ellipsoid that is above the $xy$-plane. Task 37.3 When you can use a potential to compute work, it greatly simplifies things. As you complete each problem below, first ask if there is a potential. Compute the work done by the vector field $\vec F(x,y) = (y,x)$ on a object that moves along the path $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2 \pi$. Compute the work done by the vector field $\vec F(x,y) = (-y,x)$ on a object that moves along the path $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2 \pi$. For each vector field above, use Mathematica to construct an image that shows the vector field along with the curve in the same plot. (You'll need VectorPlot[] and ParametricPlot[], with Show[] to get them in the same plot). We've seen that if a vector field has a potential, then the derivative is symmetric. Is the converse of this statement true, namely if the derivative of a vector field is symmetric, then does that mean the vector field has a potential? Consider the vector field $\ds\vec F(x,y) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$. Show that the derivative is symmetric. Compute the work done by the vector field $\vec F(x,y)$ on a object that moves along the path $\vec r(t) = (\cos t, \sin t)$ for $0\leq t\leq 2 \pi$. Explain why $\vec F(x,y)$ does not have a potential. Look up "simply connected region" (see section 6.3), and explain why the domain of $\ds\vec F(x,y) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$ is NOT simply connected. When the domain of a continuously differentiable vector field is simply connected, then the vector field has a potential if and only if the derivative is symmetric. The concept of a simply connected domain is the start of an entire branch of mathematics called algebraic topology, all stemming from the question, "under what circumstances can we guarantee that a vector field will have a potential?" Task 37.4 Pick some problems related to the topics we are discussing from the Text Book Practice page. Edit Page History Source Attach File Backlinks List Group Page last modified on November 14, 2022, at 05:05 PM
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Day 37

Task 37.1

Task 37.2

Task 37.3

Task 37.4